Active Butterworth Filter

I was given an assignment to build a circuit specified by my instructor and was told that it was a lowpass butterworth filter. However, after building the circuit in MultiSim and observing the frequency response in a Bode Plot, it looks like a bandpass filter. If you could please take a look at the circuit and the Bode plot and let me know if I am correct, or if I have made a mistake somewhere.

the filter your instructor provided is a high pass filter. you can tell this because the inputs are capacitivly coupled (ie only high frequencies can pass thru).
The circuit you drew has no feedback as already stated and would produce a high pass configuration. the pass band you are seeing in your bode plot is caused by the high pass filter created by the capacitive coupling and the open loop (dominant) pole of the op-amp.

Thank you for that information! Now I am a little confused though, what exactly is missing from my circuit?

Click to expand...

The circuit you were given should contain a connection dot at the output of the opamp. You would then need to change the connections in your simulation model to agree with your latest diagram.

The active filter schematic in your most recent attachment appears to be a high-pass rather than a low-pass topology. You could call it a bandpass in that the most common opamps do not have the ability to pass frequencies beyond a few 100 Khz. However it appears that the design was intended to operate as a second-order high-pass filter.

The circuit you were given is completely wrong:
1) It does not have negative feedback.
2) One of its resistors is shown as a capacitor so the opamp input has no bias voltage and the circuit is plain wrong.
3) The top resistor has a value that is too low for an opamp to drive.
4) The opamp is not powered.

Why not look in Google for Sallen and Key Butterworth Highpass Filter Circuit??

Your diagram in post #3 of this thread does have an error in that the capacitor from the opamp's positive input terminal to ground is labeled as 680 Ω but is draw as a capacitor. It should be a resistor not a capacitor as audioguru has stipulated. This is probably an oversight in copying the schematic into the simulation program.

Also there is the lack of a clear indication of a connection of all of the lines at the output of the opamp that I mentioned in my earlier reply to this post.

Thank you so much! The second diagram was the one given to me by my instructor and as you can see, her teaching leaves much to be desired.
The circuit now works as a high pass filter! Thank you all very much for your help!

The circuit you were given is completely wrong:
1) It does not have negative feedback.
2) One of its resistors is shown as a capacitor so the opamp input has no bias voltage and the circuit is plain wrong.
3) The top resistor has a value that is too low for an opamp to drive.
4) The opamp is not powered.

Click to expand...

This given to him is correct, except that the capacitor to ground should be a resistor (as it is designated as 680 ohms...). The device in the diagram is an ideal op-amp i believe. It configured as a high pass second order filter (if the 680 ohm "capacitor" is actually a resistor). If you look at it you see that it isa unity gain amplifier with it's first pole given by the first capacitor and the positive feedback resistor. The second pole is formed by the second capacitor and the "resistor" to ground.

A hint to achieve your goal is:

Replace the capacitors with resistors, and the resistors with capacitors. This will make a second order low pass filter with a unity gain at DC.